# Ch 8.5 Solution Concentration Units % (m/m or w/w) = mass of solute x 100 total mass of solution mass of solution = mass solute + mass solvent

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1 1 Ch 8.5 Solution Concentration Units % (m/m or w/w) = mass of solute x 100 total mass of solution mass of solution = mass solute + mass solvent % (v/v) = volume of solute x 100 volume of solution filled to a total volume of solution % (m/v or w/v) = mass of solute (g) x 100 volume of solution (ml) units are specified Parts-per-million ppm = mass or volume of solute x 10 6 mass or volume of solution

2 2 Sample calculation: How many ounces of wine (12 % v/v) are in one standard drink? 12 ml pure alcohol in 100 ml wine Density of pure alcohol = g/ml 1 ounce = ml g alcohol ml alcohol ml wine ounces wine 14 g alcohol x 1 ml x 100 ml wine x 1 oz. 1 standard drink g 12 ml alcohol ml = 5.0 oz of wine/standard drink

3 3 Molarity (M) = moles of solute L of solution Drill problem: What volume of 2.0 M CaCl 2 is needed to provide 1.94 g Ca 2+? g Ca 2+ mol Ca 2+ mol CaCl 2 volume soln 1.94 g Ca 2+ x 1 mol x 1 mol CaCl 2 x 1 L g 1 mol Ca mol CaCl 2 = L or 24 ml

4 4 Ch 8.6 Dilution C stock soln x V stock soln = C dilute soln x V dilute soln Practice problem: How much water must be added to 100 ml of M NaCl to prepare a M NaCl solution? C s x V s = C d x V d (you must know this equation) M x 100 ml = M x V d V d = 600 ml = final volume (V s + V water ) V water = V d V s = 600 ml 100 ml = 500 ml Q: How many-fold of a dilution has occurred?

5 5 Ch 8.7 Colloidal Dispersions and Suspensions Table 8.4 Property comparison for solutions, colloidal dispersions, and suspensions. Ch 8.8 Colligative Properties of Solutions reduction of vapor pressure elevation of boiling point depression of freezing point osmotic pressure directly proportional to the total concentration of all the species formed when solutes dissolve

6 6 Ch 8.9 Osmosis and Osmotic Pressure osmolarity = molarity x i i = number of particles from one formula unit of solute Example: What is the osmolarity of a solution that is 1 M in MgBr 2 and 2 M in glucose? Osmolarity = 1 M x M x 1 = 5 osmol MgBr 2 glucose The osmotic pressure of a solution is directly proportional to the number of solute particles present. Solution A 0.30 osmol = 7.6 atm Solution B 0.15 osmol = 3.8 atm Solution C 1.5 osmol =? Isotonic, Hypotonic, and Hypertonic solutions. Ch 8.10 Dialysis: a semipermeable membrane allows the passage of solvent, dissolved ions, and small molecules, but blocks the passage of larger particles. (separation technique)

7 7 Chapter 9 Chemical Reactions Ch covered in Review II (Chapter Medley) Ch 9.4 Collision Theory and Chemical Reactions Activation energy (E a ) is the minimum combined kinetic energy that colliding reactants must possess in order for their collision to result in a chemical reaction. Higher activation energy Slower reaction Lower activation energy Faster reaction Collision Orientation Reaction rates are sometimes very slow because reactant molecules must be oriented in a certain way.

8 8 Ch 9.5 Exothermic Chemical Reactions Figure 9.7 Energy diagram for an exothermic reaction. Sometimes an initial input of energy may be needed but once it has started, an exothermic reaction is selfsustaining. Ch 9.5 Endothermic Chemical Reactions Figure 9.7 Energy diagram for an endothermic reaction. A continuous input of energy is needed for endothermic reactions to occur.

9 Ch 9.6 Factors That Influence Chemical Reaction Rates Physical nature of the reactants Reactant concentrations (surface area for solids) Reaction temperature (rate doubles every 10 o C) Reaction pressure in the case of gases Presence of catalysts Catalyst is not consumed. It provides an alternative route for the reaction with a lower activation energy (E a ). Catalysts DO NOT influence the amount of product formed, they only speed up the process. 9

10 10 Drill Problem: Will each of the changes listed increase or decrease the rate of the following reaction? N H 2 2 NH 3 Adding some N 2 Raising T Removing a catalyst Removing some H 2 Drill Problem: Reaction A H = -20 kcal/mole E a = 25 kcal/mole Reaction B H = -15 kcal/mole E a = 20 kcal/mole Which reaction occurs faster at the same temperature? Which reaction releases more heat energy?

11 11 Ch 9.7 Chemical Equilibrium = Dynamic Equilibrium Ch 9.8 Equilibrium Constant wa + xb yc + zd [products] K eq = = [reactants] [ ] = molar concentrations (M = moles/l) [C] y [D] z [A] w [B] x When [products] >> [reactants], K eq = large, Equilibrium right When [reactants] >> [products], K eq = small, Equilibrium left At values of K eq > 1000, [products] >> [reactants] and the reaction is usually considered complete. Pure solids and pure liquids have constant concentrations, which are incorporated into the equilibrium constant itself. 2 KClO 3 (s) 2 KCl(s) + 3 O 2 (g) K eq = [O 2 ] 3

12 12 N 2 (g) + 3 H 2 (g) catalyst 2 NH 3 (g) Sample calculation: K eq = 70 at 350 o C [N 2 ] = M [H 2 ] = M [NH 3 ] =? [NH 3 ] 2 [NH 3 ] 2 K eq = = = 70 [N 2 ][H 2 ] 3 [0.100][0.300] 3 [NH 3 ] 2 = 70 x x (0.300) 3 [NH 3 ] 2 = [NH 3 ] 2 = [NH 3 ] = M

13 13 Ch 9.9 Altering Equilibrium Conditions Concentration changes (K eq is unchanged) Pressure changes (K eq is unchanged) Temperature changes (K eq changes) Addition of catalyst has NO EFFECT on the equilibrium position, it merely allows the equilibrium to be reached more quickly. (K eq is unchanged) N 2 (g) + 3 H 2 (g) Stress imposed Add N 2 Remove N 2 Add H 2 Remove H 2 Add NH 3 Remove NH 3 2 NH 3 (g) Shift observed right left right left left right

14 14 Change in pressure (K eq is unchanged). Drill Problem: Rxn A: 2 NO 2 (g) + 7 H 2 (g) 2 NH 3 (g) + 4 H 2 O(g) Rxn B: H 2 (g) + I 2 (g) 2 HI(g) What will be the effects of increasing pressure? Rxn A: Rxn B: What will be the effects of decreasing pressure? Rxn A: Rxn B:

15 15 When the reaction temperature changes, K eq also changes! H 2 (g) + F 2 (g) 2 HF(g) + Heat For exothermic reactions heat = product Increase T, shift left - [HF] decreases, K eq decreases Decrease T, shift right - [HF] increases, K eq increases 2 CO 2 (g) + Heat 2 CO(g) + O 2 (g) For endothermic reactions heat = reactant Increase T, shift right - K eq increases Decrease T, shift left - K eq decreases

16 16 Supplemental material: Spontaneity and Chemical Reactions Enthaply ( H) and Entropy ( S) determine whether a reaction is spontaneous or nonspontaneous Most spontaneous chemical reactions are exothermic. Many endothermic reactions are nonspontaneous. Some endothermic reactions are spontaneous because Entropy increases: S = positive

17 17 Drill Problem: PCl 5 (g) PCl 3 (g) + Cl 2 (g) H = (+) 1. What changes would produce more PCl 5 (g)? [ ]? add Cl 2 or PCl 3 P? increase P to shift equilibrium left T? H = (+) for endothermic reaction and Heat is on left side, decrease T to favor left side, K eq changes Catalyst? No effect on equilibrium, just reaction rate. 2. What is the sign of S in the forward direction? 1 mole of gas 2 moles of gas = disorder increases, S = + 3. A 4.0 L flask at equilibrium contains 0.60 mole PCl mole PCl mole Cl 2 Calculate the K eq for the reaction. K eq = [PCl 3 ][Cl 2 ] = (0.20mol/4L)(1.0mol/4L) = [PCl 5 ] (0.60mol/4L)

18 Chapter 10 Acids, Bases, and Salts Ch 10.1 Arrhenius Acid-Base Theory (also in Chapter Medley) Arrhenius Acids produce H + in water Arrhenius Bases produce OH - in water HCl hydrochloric acid KOH HNO 3 nitric acid Ba(OH) 2 HClO 4 perchloric acid H 2 SO 4 sulfuric acid H 3 PO 4 phosphoric acid 18

19 19 Ch 10.2 Brønsted Lowry Acid-Base Theory Brønsted Lowry acid = proton (H + ion) donor Brønsted Lowry base = proton (H + ion) acceptor Hydronium Ion Base H + acceptor Acid H + donor H 2 O(l) + HCl(g) H 3 O + (aq) + Cl - (aq) Only acidic H atoms are donated: Acetic acid is monoprotic Acidic Substances that can either donate or accept a H + are called amphiprotic.

20 20 Conjugate Acid-Base pairs differ by a single H + Drill Problem. Write the chemical formula for: - 1. The conjugate base of H 2 PO 4-2. The conjugate acid of H 2 PO 4 3. The conjugate base of H 2 O 4. The conjugate acid of H 2 O

21 21 Ch 10.3 Mono-, Di-, and Triprotic Acids Monoprotic acids can transfer 1 H + to H 2 O or base Examples: HCl and HNO 3 Diprotic acids can transfer 2 H + Example: H 2 SO 4 + H 2 O H 3 O + + HSO 4 - HSO H 2 O H 3 O + + SO 4 2- Triprotic acids can transfer 3 H + Example: H 3 PO 4 + H 2 O H 2 PO 4 + H 2 O 2 HPO 4 + H 2 O H 3 O + + H 2 PO 4 H 3 O HPO 4 H 3 O PO 4 A polyprotic acid supplies 2 or more H +

22 Ch 10.4 Strengths of Acids and Bases A strong acid donates all or nearly 100% of its H + to H 2 O 22 Table 10.1 Learn the names and formulas of these commonly encountered strong acids, and then assume that all other acids you encounter are weak, unless you are told otherwise. These acids are strong even in dilute solution because in water they are all or mostly ionized. A weak acid does not ionize completely. CH 3 CO 2 H(l) + H 2 O(l) CH 3 CO 2 - (aq) + H 3 O + (aq)

23 23 Strong bases are limited to the hydroxides of Group IA and IIA listed in Table Ammonia gas (NH 3 ) is the most common weak base. NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq)

24 24 Ch 10.5 Ionization Constants for Acids and Bases HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) acid ionization constant K a = [H 3 O + ][A - ] [HA] Acid strength increases with increasing K a values. B(aq) + H 2 O(l) BH + (aq) + OH - (aq) base ionization constant K b = [BH + ][OH - ] [B] Base strength increases with increasing K b values.

25 25 Acid-Base Neutralization Reactions (Ch 10.1, 10.6 & 10.7 covered in Chapter Medley) Acid + Base Salt + Water HX + BOH BX + HOH Net ionic equation: H + + OH - H 2 O Ch 10.8 Self-Ionization of Water H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) K w = [H 3 O + ][OH - ] = 1.00 x memorize this # K w = Ion product constant for pure H 2 O at 25 o C Pure H 2 O at 25 o C, [H 3 O + ] = [OH - ] = 1.00 x 10-7 M Neutral

26 26 HCl is added to produce a M solution: M HCl M H 3 O + acidic solution [H 3 O + ] = 1.0 x 10-2 acidic solution [H 3 O + ] > 10-7 and [H 3 O + ] > [OH - ] K w = [H 3 O + ][OH - ] = 1.00 x [OH - ] = K w /[H 3 O + ] = 1.00 x /1.0 x 10-2 [OH - ] = 1.0 x NaOH is added to produce a M solution: M NaOH M OH - basic solution [OH - ] = 1.0 x 10-3 basic solution [OH - ] > 10-7 and [OH - ] > [H 3 O + ] K w = [H 3 O + ][OH - ] = 1.00 x [H 3 O + ] = K w /[OH - ] = 1.00 x /1.0 x 10-3 [H 3 O + ] = 1.0 x 10-11

27 27 Ch 10.9 The ph concept ph is the negative logarithm of the molarity of the hydronium ion: ph = -log[h 3 O + ] [H O + ] = 6.3 x ph = sig fig 2 digits enter 6.3 x 10-5 in your calculator, press the log key, then switch the sign ph = 7 = neutral ph < 7 = acidic ph > 7 = basic

28 28 Ch The pk a Method for Expressing Acid Strength pk a = -logk a Acid strength increases with increasing K a increases with decreasing pk a Drill question: Which is the stronger acid? acetic acid K a = 1.8 x 10-5 pk a = 4.74 HF K a = 6.8 x 10-4 pk a = 3.17 Ch The ph of Aqueous Salt Solutions Table 10.6 Examples of neutral, acidic, and basic salts.

29 29 Ch Buffers a weak acid and its conjugate base CH 3 CO 2 H/CH 3 CO 2 - H 2 PO 4 - /HPO 4 2- H 2 CO 3 /HCO 3 - (buffer in human blood) Added base is absorbed by the acid OH - + H 2 CO 3 HCO H 2 O Added acid is absorbed by the conjugate base H 3 O + + HCO 3 - H 2 CO 3 + H 2 O

30 Ch Electrolytes Strong Electrolyte = soluble ionic compounds and covalent molecules that ionize completely, such as strong acids Weak electrolytes = molecules that ionize incompletely, such as weak acids and bases Nonelectrolyte = covalent molecules that do not ionize Drill Problem. Classify each of the following compounds as a strong electrolyte, weak electrolyte, or nonelectrolyte: H 3 PO 4 HCl Cl 2 HF KBr CH 3 CH 2 -OH CH 3 COOH 30

31 31 Ch Equivalents and Milliequivalents of Electrolytes One equivalent (Eq) supplies one mole of charge 1 mole Na + = 1 equivalent 1 mole Ca 2+ = 2 equivalents 3-1 mole PO 4 = 3 equivalents Sample Calculation. Human blood plasma contains 2.4 mg Mg 2+ per dl. How many Eq or meq are in 1.0 L of plasma? Strategy: dl L mg g mol Eq meq 1.0 L x 2.4 mg Mg 2+ x 10 dl x. 1 g. x 1 mol x 2 Eq Mg 2+ 1 dl 1 L 10 3 mg 24.3 g mol Mg 2+ = 2.0 x 10-3 Eq Mg 2+ or 2.0 meq Mg 2+

32 32 Ch Acid-Base Titrations endpoint detected with acid-base indicator: HInd + H 2 O H 3 O + + Ind - Sample calculation. In an acid-base titration, ml of M H 3 PO 4 is needed to neutralize 25.0 ml of KOH of unknown concentration. Calculate the molarity of the KOH. H 3 PO 4 (aq) + 3 KOH(aq) K 3 PO 4 (aq) + 3 H 2 O(l) ml x mol H 3 PO 4 x 3 mole KOH = mole KOH 1000 ml 1 mol H 3 PO 4 M = mol/l = mole KOH/0.025 L = M KOH

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